356205
In an ac circuit, the current is given by \({i=5 \sin (100 t-\pi / 2)}\) and the ac potential is \({V=200 \sin (100) t}\), the power consumption is
1 \(20\,W\)
2 \(40\,W\)
3 \(100\,W\)
4 \(0\,W\)
Explanation:
Phase angle between \({V}\) and \({I}\) is \({90^{\circ}}\). \( < P > = {V_{{\rm{rms }}}}{I_{{\rm{rms }}}}\cos \frac{\pi }{2} = 0\)
PHXII07:ALTERNATING CURRENT
356206
In an \(A.C.\) circuit \(V\) and \(I\) are given by \(V = 150\sin (150t)\) volt and \(I = 150\sin (150t + \frac{\pi }{3})\) ampere. The power dissipated in the circuit is
1 \(106\,W\)
2 \(150\,W\)
3 \(5625\,W\)
4 \({\rm{zero}}\)
Explanation:
Comparing \(V = 150\sin (150t)\) with \(V = {V_0}\sin \omega t,\) we get \({V_0} = 150V\) Comparing \(I = 150\sin \left( {150t + \frac{\pi }{3}} \right)\) with \(I = {I_0}\sin (\omega t + \phi )\) we get \({I_0} = 150A,\phi = \frac{\pi }{3} = 60^\circ \) The power dissipated in \(A.C\) circuit is \(P = \frac{1}{2}{V_0}{I_0}\cos \phi = \frac{1}{2} \times 150 \times 150 \times \cos 60^\circ \) \( = \frac{1}{2} \times 150 \times 150 \times \frac{1}{2} = 5265\,W\)
PHXII07:ALTERNATING CURRENT
356207
The power factor in a circuit connected to an AC power supply has a value which is
1 unity when the circuit contains only inductance
2 unity when the circuit contains only resistance
3 zero when the circuit contains an ideal resistance only
4 unity when the circuit contains an ideal capacitance only
Explanation:
Power factor, \(\cos \phi = \frac{R}{Z}\) When circuit contains only resistance, then \(Z = R\) \(\therefore \,\,\,\,\,\,\,\cos \phi = \frac{R}{R} = 1\)
PHXII07:ALTERNATING CURRENT
356208
An alternating voltage \(V = {V_0}\sin \omega t\) is applied across a circuit. As a result, a current \(I = {I_0}\sin (\omega t - \pi /2)\) flows in it. The power consumed per cycle is
1 zero
2 \(0.5\;{V_0}{I_0}\)
3 \(0.707\;{V_0}{I_0}\)
4 \(1.414\;{V_0}{I_0}\)
Explanation:
The phase angle between voltage \(V\) and current \(I\) is \(\pi / 2\).
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PHXII07:ALTERNATING CURRENT
356205
In an ac circuit, the current is given by \({i=5 \sin (100 t-\pi / 2)}\) and the ac potential is \({V=200 \sin (100) t}\), the power consumption is
1 \(20\,W\)
2 \(40\,W\)
3 \(100\,W\)
4 \(0\,W\)
Explanation:
Phase angle between \({V}\) and \({I}\) is \({90^{\circ}}\). \( < P > = {V_{{\rm{rms }}}}{I_{{\rm{rms }}}}\cos \frac{\pi }{2} = 0\)
PHXII07:ALTERNATING CURRENT
356206
In an \(A.C.\) circuit \(V\) and \(I\) are given by \(V = 150\sin (150t)\) volt and \(I = 150\sin (150t + \frac{\pi }{3})\) ampere. The power dissipated in the circuit is
1 \(106\,W\)
2 \(150\,W\)
3 \(5625\,W\)
4 \({\rm{zero}}\)
Explanation:
Comparing \(V = 150\sin (150t)\) with \(V = {V_0}\sin \omega t,\) we get \({V_0} = 150V\) Comparing \(I = 150\sin \left( {150t + \frac{\pi }{3}} \right)\) with \(I = {I_0}\sin (\omega t + \phi )\) we get \({I_0} = 150A,\phi = \frac{\pi }{3} = 60^\circ \) The power dissipated in \(A.C\) circuit is \(P = \frac{1}{2}{V_0}{I_0}\cos \phi = \frac{1}{2} \times 150 \times 150 \times \cos 60^\circ \) \( = \frac{1}{2} \times 150 \times 150 \times \frac{1}{2} = 5265\,W\)
PHXII07:ALTERNATING CURRENT
356207
The power factor in a circuit connected to an AC power supply has a value which is
1 unity when the circuit contains only inductance
2 unity when the circuit contains only resistance
3 zero when the circuit contains an ideal resistance only
4 unity when the circuit contains an ideal capacitance only
Explanation:
Power factor, \(\cos \phi = \frac{R}{Z}\) When circuit contains only resistance, then \(Z = R\) \(\therefore \,\,\,\,\,\,\,\cos \phi = \frac{R}{R} = 1\)
PHXII07:ALTERNATING CURRENT
356208
An alternating voltage \(V = {V_0}\sin \omega t\) is applied across a circuit. As a result, a current \(I = {I_0}\sin (\omega t - \pi /2)\) flows in it. The power consumed per cycle is
1 zero
2 \(0.5\;{V_0}{I_0}\)
3 \(0.707\;{V_0}{I_0}\)
4 \(1.414\;{V_0}{I_0}\)
Explanation:
The phase angle between voltage \(V\) and current \(I\) is \(\pi / 2\).
356205
In an ac circuit, the current is given by \({i=5 \sin (100 t-\pi / 2)}\) and the ac potential is \({V=200 \sin (100) t}\), the power consumption is
1 \(20\,W\)
2 \(40\,W\)
3 \(100\,W\)
4 \(0\,W\)
Explanation:
Phase angle between \({V}\) and \({I}\) is \({90^{\circ}}\). \( < P > = {V_{{\rm{rms }}}}{I_{{\rm{rms }}}}\cos \frac{\pi }{2} = 0\)
PHXII07:ALTERNATING CURRENT
356206
In an \(A.C.\) circuit \(V\) and \(I\) are given by \(V = 150\sin (150t)\) volt and \(I = 150\sin (150t + \frac{\pi }{3})\) ampere. The power dissipated in the circuit is
1 \(106\,W\)
2 \(150\,W\)
3 \(5625\,W\)
4 \({\rm{zero}}\)
Explanation:
Comparing \(V = 150\sin (150t)\) with \(V = {V_0}\sin \omega t,\) we get \({V_0} = 150V\) Comparing \(I = 150\sin \left( {150t + \frac{\pi }{3}} \right)\) with \(I = {I_0}\sin (\omega t + \phi )\) we get \({I_0} = 150A,\phi = \frac{\pi }{3} = 60^\circ \) The power dissipated in \(A.C\) circuit is \(P = \frac{1}{2}{V_0}{I_0}\cos \phi = \frac{1}{2} \times 150 \times 150 \times \cos 60^\circ \) \( = \frac{1}{2} \times 150 \times 150 \times \frac{1}{2} = 5265\,W\)
PHXII07:ALTERNATING CURRENT
356207
The power factor in a circuit connected to an AC power supply has a value which is
1 unity when the circuit contains only inductance
2 unity when the circuit contains only resistance
3 zero when the circuit contains an ideal resistance only
4 unity when the circuit contains an ideal capacitance only
Explanation:
Power factor, \(\cos \phi = \frac{R}{Z}\) When circuit contains only resistance, then \(Z = R\) \(\therefore \,\,\,\,\,\,\,\cos \phi = \frac{R}{R} = 1\)
PHXII07:ALTERNATING CURRENT
356208
An alternating voltage \(V = {V_0}\sin \omega t\) is applied across a circuit. As a result, a current \(I = {I_0}\sin (\omega t - \pi /2)\) flows in it. The power consumed per cycle is
1 zero
2 \(0.5\;{V_0}{I_0}\)
3 \(0.707\;{V_0}{I_0}\)
4 \(1.414\;{V_0}{I_0}\)
Explanation:
The phase angle between voltage \(V\) and current \(I\) is \(\pi / 2\).
356205
In an ac circuit, the current is given by \({i=5 \sin (100 t-\pi / 2)}\) and the ac potential is \({V=200 \sin (100) t}\), the power consumption is
1 \(20\,W\)
2 \(40\,W\)
3 \(100\,W\)
4 \(0\,W\)
Explanation:
Phase angle between \({V}\) and \({I}\) is \({90^{\circ}}\). \( < P > = {V_{{\rm{rms }}}}{I_{{\rm{rms }}}}\cos \frac{\pi }{2} = 0\)
PHXII07:ALTERNATING CURRENT
356206
In an \(A.C.\) circuit \(V\) and \(I\) are given by \(V = 150\sin (150t)\) volt and \(I = 150\sin (150t + \frac{\pi }{3})\) ampere. The power dissipated in the circuit is
1 \(106\,W\)
2 \(150\,W\)
3 \(5625\,W\)
4 \({\rm{zero}}\)
Explanation:
Comparing \(V = 150\sin (150t)\) with \(V = {V_0}\sin \omega t,\) we get \({V_0} = 150V\) Comparing \(I = 150\sin \left( {150t + \frac{\pi }{3}} \right)\) with \(I = {I_0}\sin (\omega t + \phi )\) we get \({I_0} = 150A,\phi = \frac{\pi }{3} = 60^\circ \) The power dissipated in \(A.C\) circuit is \(P = \frac{1}{2}{V_0}{I_0}\cos \phi = \frac{1}{2} \times 150 \times 150 \times \cos 60^\circ \) \( = \frac{1}{2} \times 150 \times 150 \times \frac{1}{2} = 5265\,W\)
PHXII07:ALTERNATING CURRENT
356207
The power factor in a circuit connected to an AC power supply has a value which is
1 unity when the circuit contains only inductance
2 unity when the circuit contains only resistance
3 zero when the circuit contains an ideal resistance only
4 unity when the circuit contains an ideal capacitance only
Explanation:
Power factor, \(\cos \phi = \frac{R}{Z}\) When circuit contains only resistance, then \(Z = R\) \(\therefore \,\,\,\,\,\,\,\cos \phi = \frac{R}{R} = 1\)
PHXII07:ALTERNATING CURRENT
356208
An alternating voltage \(V = {V_0}\sin \omega t\) is applied across a circuit. As a result, a current \(I = {I_0}\sin (\omega t - \pi /2)\) flows in it. The power consumed per cycle is
1 zero
2 \(0.5\;{V_0}{I_0}\)
3 \(0.707\;{V_0}{I_0}\)
4 \(1.414\;{V_0}{I_0}\)
Explanation:
The phase angle between voltage \(V\) and current \(I\) is \(\pi / 2\).